Context My question is about the matrix dimension of the shape operator.
In order to avoid misunderstanding let $S \subset \mathbb{R}^3$ be a regular surface and $$\psi(u,v)=(x(u,v),y(u,v),z(u,v))$$ be a chart on a open subset $W\subset S$. So, we have the induced basis of $T_pM$ given by $\left\{\frac{\partial \psi}{\partial u},\frac{\partial \psi}{\partial v}\right\}$ which allow us to define the Gauss map
$$p \mapsto N(u,v)=\frac{\frac{\partial \psi}{\partial u}\land \frac{\partial \psi}{\partial v}}{\big\vert\big\vert \frac{\partial \psi}{\partial u}\land \frac{\partial \psi}{\partial v}\big\vert\big\vert}$$
Now we can finally define the shape operator $S:\mathfrak X(W)\rightarrow \mathfrak X(W)$; $S(X)=- D_XN$
Question
The matrix that represents the linear operator $D_XN$ lies on the space of $3\times 2$ matrices (here we have a very good derivation about this fact), so by definition of $S$, its matrix representation should also lies on the space of $3\times 2$ matrices.
But at the same time when I'm looking on textbooks I only find that this matrix is a $2\times 2$ matrix whose entries are given by the the one and second fundamental forms coefficients (here gives the formula I'm talking about and here we have an example).
How can I should understand whats going on and how can I interpret this conceptual difference envolved ?
Thank you in advance for any hint about this question :)